I am starting on a Phd program and am supposed to read Colliot Thelene and Sansuc's article
on R-equivalence for tori. I find it very difficult and although I have some knowledge over schemes , I am completely baffled by this scalar restriction business of having a field extension $K/k$ , a torus over $K$ and "restricting" it to $k$.
I would be very gratefull for a reference or even better by some explanation . I found nothing in my standard books (Hartshorne, Qing Liu, Mumford etc) so I hope this question is appropriate for the site. Thank you.

3 Answers
3

As said, the sought after concept is also known as Weil restriction. In a word, it is the algebraic analogue of the process of viewing an $n$-dimensional complex variety as a $(2n)$-dimensional real variety.

The setup is as follows: let $L/K$ be a finite degree field extension and let $X$ be a scheme over $L$. Then the Weil restriction $W_{L/K} X$ is the $K$-scheme representing the following functor on the category of K-algebras:

$A\mapsto X(A \otimes_K L)$.

In particular, one has $W_{L/K} X(K) = X(L)$.

By abstract nonsense (Yoneda...), if such a scheme exists it is uniquely determined by the above functor. For existence, some hypotheses are necessary, but I believe that it exists whenever $X$ is reduced of finite type.

Now for a more concrete description. Suppose $X = \mathrm{Spec} L[y_1,...,y_n]/J$ is an affine scheme. Let $d = [L:K]$ and $a_1,...,a_d$ be a $K$-basis of $L$. Then we make the following "substitution":

$$y_i = a_1 x_{i1} + ... + a_d x_{id},$$

thus replacing each $y_i$ by a linear expression in d new variables $x_{ij}$. Moreover,
suppose $J = \langle g_1,...,g_m \rangle$; then we substitute each of the above equations into $g_k(y_1,...,y_n)$ getting a polynomial in the $x$-variables, however still with $L$-coefficients. But now using our fixed basis of $L/K$, we can regard a single polynomial with $L$-coefficients as a vector of $d$ polynomials with $K$ coefficients. Thus we end up with $md$ generating polynomials in the $x$-variables, say generating an ideal $I$ in $K[x_{ij}]$, and we put $\mathrm Res_{L/K} X = \mathrm{Spec} K[x_{ij}]/I$.

A great example to look at is the case $X = G_m$ (multiplicative group) over $L = \mathbb{C}$ (complex numbers) and $K = \mathbb{R}$. Then $X$ is the spectrum of

$$\mathbb{C}[y_1,y_2]/(y_1 y_2 - 1);$$

put $y_i = x_{i1} + \sqrt{-1} x_{i2}$ and do the algebra. You can really see that the
corresponding real affine variety is $\mathbb{R}\left[x,y\right]\left[(x^2+y^2)^{-1}\right]$, as it should be: see e.g.
p. 2 of

Note the important general property that for a variety $X/L$, the dimension of the Weil restriction from $L$ down to $K$ is $[L:K]$ times the dimension of $X/L$. This is good to keep in mind so as not to confuse it with another possible interpretation of "restriction of scalars", namely composition of the map $X \to \mathrm{Spec} L$ with the map $\mathrm{Spec} L \to Spec K$ to
give a map $X \to \mathrm{Spec} K$. This is a much weirder functor, which preserves the dimension but screws up things like geometric integrality. (When I first heard about "restriction of
scalars", I guessed it was this latter thing and got very confused.)

Thanks for this long answer and the link to your notes.
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user2292Dec 4 '09 at 21:41

1

Pete: While this is generally a great answer, the restriction of scalars of (G_m)_C is not A^2_R - {0,0}. Rather, it is Spec R[x,y][ (x^2+y^2)^{-1} ]. These have the same real points, but they are not the same scheme!
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David SpeyerDec 6 '09 at 3:39

@David: you're right. In the back of my mind I was wondering about how the Weil restriction of an affine variety became quasi-affine. I'll change it.
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Pete L. ClarkDec 6 '09 at 6:07

There is another description for tori. The category of tori over a field F is equivalent to the category of finite-dimensional $G_F$-lattices. Now, there is an operation of induction for group representations that converts a $G_K$-lattice into a $G_k$-lattice; this is the lattice you need.

Giving the points over one field of a torus doesn't even begin to describe it. the comment "it's not hard at all" is useless and patronising.
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user2292Dec 5 '09 at 21:36

As other people commented, for a definition and properties you could check out wikipedia first. I gave you an example. And I do believe that this kind of stuff is easy.
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Evgeny ShinderDec 29 '09 at 5:15

1

I think Evgeny's example is a good one to keep in mind. (As Pete Clark wrote in his answer, restriction of scalars is just the process of thinking of a ${\mathbb C}$-variety, say, as an ${\mathbb R}$-variety instead.)
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EmertonJan 16 '10 at 4:33